Answer:
- number of multiplies is n!
- n=10, 3.6 ms
- n=15, 21.8 min
- n=20, 77.09 yr
- n=25, 4.9×10^8 yr
Step-by-step explanation:
Expansion of a 2×2 determinant requires 2 multiplications. Expansion of an n×n determinant multiplies each of the n elements of a row or column by its (n-1)×(n-1) cofactor determinant. Then the number of multiplies is ...
mpy[n] = n·mp[n-1]
mpy[2] = 2
So, ...
mpy[n] = n! . . . n ≥ 2
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If each multiplication takes 1 nanosecond, then a 10×10 matrix requires ...
10! × 10^-9 s ≈ 0.0036288 s ≈ 0.004 s . . . for 10×10
Then the larger matrices take ...
n=15, 15! × 10^-9 ≈ 1307.67 s ≈ 21.8 min
n=20, 20! × 10^-9 ≈ 2.4329×10^9 s ≈ 77.09 years
n=25, 25! × 10^-9 ≈ 1.55112×10^16 s ≈ 4.915×10^8 years
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For the shorter time periods (less than 100 years), we use 365.25 days per year.
For the longer time periods (more than 400 years), we use 365.2425 days per year.
x - Ram's age now
Ram is half his father's age now, so his father's age is two times Ram's age:
2x - Ram's father age now
x - 10 - Ram's age ten years ago
2x - 10 - Ram's father age ten years ago
ten years ago he was one third his father's age so his father's age was three times Ram's age:
2x - 10 = 3•(x - 10)
2x - 10 = 3x - 30
-10 + 30 = 3x - 2x
20 = x
2x = 2•20 = 40
Answer:
<u>Ram is 20 years old now, Ram's father is 40 years old now.</u>
Answer:
Step-by-step explanation:
Answer: 2z-15
Step-by-step explanation
Based on the visual image, AD is a line where a point O intersects somewhere on the line. AD is represented as 3z-11, while the segment from AO is represented as z+4.
In order to find the answer, we must subtract the entire line (AD) from AO in order to find the missing segment OD.
AD - AO = OD
(3z-11) - (z+4) = OD
Distribute the negative sign so it becomes:
(3z-11) -z - 4 = OD
Now combine like terms.
(3z-z) +(- 11 -4)= OD
2z - 15 = OD
And that is your answer!
